2 Marks Questions

Question 12 Marks

Find the length of perpendicular from the origin to the lines $7x + 24y = 50.$

Answer

Here, it is given: Point $(0,0)$ and line $7x + 24y = 50$
We have to find: The length of the perpendicular from the origin to the line $7x + 24y = 50$
We know that the length of the perpendicular from $P (m,n)$ to the line $ax + by + c = 0$ is given by,

The given equation of the line is $7x + 24y - 50=0 $
Here $m= 0$ and $n = 0, a = 7, b = 24, c = -50$
$D=\frac{|7(0)+24(0)-50|}{\sqrt{7^2+24^2}}$
$D=\frac{|0+0-50|}{\sqrt{49+576}}=\frac{|-50|}{\sqrt{625}}=\frac{|-50|}{25}=\frac{50}{25}=2$
$D =2$
Therefore, the length of perpendicular from the origin to the line $7x + 24y = 50$ is $2$ units.

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Question 22 Marks

In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

Answer

Let H be the set of students who know Hindi and E be the set of students who know English.
Here $n ( H )=100, n ( E )=50$ and $n(H \cap E)= 2 5$
We know that $n(H \cup E)=n(H)+n(E)-n(H \cap E)$

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Question 32 Marks

Check whether the probabilities $P ( A )$ and $P ( B )$ are consistently defined $P ( A )=0.5, P ( B )=0.4, P(A \cup B)=0.8$

Answer

Given that $P ( A )=0.5, P ( B )=0.4$ and $P(A \cup B)=0.8$
Applying the general addition rule,
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\therefore 0.8=0.5+0.4-P(A \cap B)$
$\Rightarrow P(A \cap B)=0.9-0.8=0.1$
$\therefore P(A \cap B)<P(A)$ and $P(A \cap B)<P(B)$
Thus the given probabilities are consistently defined.

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Question 42 Marks

If $A$ and $B$ are two events associated with a random experiment such that $P(A) = 0.25, P(B) = 0.4$ and $P(A$ or $B) = 0.5,$ find the values of
$i. P ( A$ and $B )$
$ii. P ( A$ and $\bar{B})$

Answer

$i$. It is given that
$: P ( A )=0.25, P ( A$ or $B )=0.5$ and $P ( B )=0.4$
To find : $P(A$ and $B)$
Formula used : $P(A$ or $B) = P(A) + P(B) - P(A $ and $B)$
Substituting the value in the above formula we get,
$0.5 = 0.25 + 0.4 - P(A$ and $B)$
$0.5 = 0.65 - P(A$ and $B)$
$P(A$ and $B) = 0.65 - 0.5$
$P(A$ and $B) = 0.15$
$ii.$ Given :$ P(A) = 0.25, P(A$ and $B) = 0.15\ ($ from part $(i))$
To find: $P ( A$ and $\bar{B})$
Formula used: $P ( A$ and $\vec{B})= P ( A )- P ( A$ and $B )$
Substituting the value in the above formula we get,
$ P ( A $ and $ \vec{B})=0.25-0.15$
$P ( A $ and $ \bar{B})=0.10$
$P ( A $ and $ \bar{B})=0.10$

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Question 52 Marks

Evaluate $\lim _{x \rightarrow 1} \frac{x^3-1}{x-1}$

Answer

Given, $\lim _{x \rightarrow 1} \frac{x^3-1}{x-1}$
If we put $x=1$, then expression $\frac{x^3-1}{x-1}$ becomes the indeterminate form $\frac{0}{0}$.
Therefore, $(x-1)$ is a common factor of $\left(x^3-1\right)$ and $(x-1)$.
Factorising the numerator and denominator, we have
$\lim _{x \rightarrow 1} \frac{x^3-1}{x-1}\left[\frac{0}{0}\right.$ form $]$
$=\lim _{x \rightarrow 1} \frac{(x-1)\left(x^2+x+1\right)}{(x-1)}$
$=\lim _{x \rightarrow 1}\left(x^2+x+1\right)$
$=1^2+1+1$
$=3$

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Question 62 Marks

Find the domain and the range of the real function: $f ( x )=\frac{|x-4|}{x-4}$

Answer

Here we have, $f(x)=\frac{|x-4|}{x-4}$
We need to find where the function is defined.
To find the domain of the function f(x) we need to equate the denominator of the function to 0
Therefore,
x - 4 = 0 or x = 4
It means that the denominator is zero when x = 4
So, the domain of the function is the set of all the real numbers except 4
The domain of the function, $D _{\{f( x )\}}=(-\infty, 4) \cup(4, \infty)$
The numerator is an absolute function of the denominator.
So, for any value of x from the domain set, we always get either +1 or -1 as the output.
So, the range of the function is a set containing -1 and +1
Therefore, the range of the function, $R _{ f ( x )}=\{-1,1\}$

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Question 72 Marks

Let $f , g$ be two real functions defined by $f(x)=\sqrt{x+1}$ and $g(x)=\sqrt{9-x^2}$ Then describe each of the following functions: $f + g$.

Answer

According to the question , we can state,
We know the square of a real number is never negative.
Clearly, $f ( x )$ takes real values only when $x+1 \geq 0$
$= x > -1$
$\therefore x \in[-1, \infty)$
Thus, domain of $f=(-1, \infty)$
Similarly, $g(x)$ takes real values only when $9-x^2 \geq 0$
$=9>x^2$
$=x^2<9$
$=x^2-9<0$
$=x^2-32<0$
$=(x+3)(x-3)<0$
$=x \geq-3$ and $x<3$
$x \in[-3,3]$
Thus, domain of $g==[-3,3]$
$i.f + g$
We know $(f + g)(x) = f(x) + g(x)$
$\therefore(f+g)(x)=\sqrt{x+1}+\sqrt{9-x^2}$
Domain of $f+g=$ Domain of $f \cap$ Domain of $g$
$=$ Domain of $f+g=[-1, \infty) \cap[-3,3]$
Thus, $f + g :[-1,3] R$ is given by $(f+g)(x)$
$=\sqrt{x+1}+\sqrt{9-x^2}$

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